Problem: Alice has a cute cat. To keep her cat fit, Alice wants to design an exercising walk for her cat! 

Initially, Alice's cat is located in a cell (x,y) of an infinite grid. According to Alice's theory, cat needs to move: 

  * exactly a steps left: from (u,v) to (u-1,v); 
  * exactly b steps right: from (u,v) to (u+1,v); 
  * exactly c steps down: from (u,v) to (u,v-1); 
  * exactly d steps up: from (u,v) to (u,v+1). 



Note that the moves can be performed in an arbitrary order. For example, if the cat has to move 1 step left, 3 steps right and 2 steps down, then the walk right, down, left, right, right, down is valid.

Alice, however, is worrying that her cat might get lost if it moves far away from her. So she hopes that her cat is always in the area [x_1,x_2]× [y_1,y_2], i.e. for every cat's position (u,v) of a walk x_1 ≤ u ≤ x_2 and y_1 ≤ v ≤ y_2 holds.

Also, note that the cat can visit the same cell multiple times.

Can you help Alice find out if there exists a walk satisfying her wishes?

Formally, the walk should contain exactly a+b+c+d unit moves (a to the left, b to the right, c to the down, d to the up). Alice can do the moves in any order. Her current position (u, v) should always satisfy the constraints: x_1 ≤ u ≤ x_2, y_1 ≤ v ≤ y_2. The staring point is (x, y).

You are required to answer t test cases independently.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^3) — the number of testcases. 

The first line of each test case contains four integers a, b, c, d (0 ≤ a,b,c,d ≤ 10^8, a+b+c+d ≥ 1).

The second line of the test case contains six integers x, y, x_1, y_1, x_2, y_2 (-10^8 ≤ x_1≤ x ≤ x_2 ≤ 10^8, -10^8 ≤ y_1 ≤ y ≤ y_2 ≤ 10^8).

Output

For each test case, output "YES" in a separate line, if there exists a walk satisfying her wishes. Otherwise, output "NO" in a separate line. 

You can print each letter in any case (upper or lower).

Example

Input


6
3 2 2 2
0 0 -2 -2 2 2
3 1 4 1
0 0 -1 -1 1 1
1 1 1 1
1 1 1 1 1 1
0 0 0 1
0 0 0 0 0 1
5 1 1 1
0 0 -100 -100 0 100
1 1 5 1
0 0 -100 -100 100 0


Output


Yes
No
No
Yes
Yes
Yes

Note

In the first test case, one valid exercising walk is $$$(0,0)→ (-1,0) → (-2,0)→ (-2,1) → (-2,2)→ (-1,2)→(0,2)→ (0,1)→ (0,0) → (-1,0)$$$

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("O3")
using namespace std;
long long mul(long long a, long long b, long long m = 1000000007) {
  long long res = 0;
  a = a % m;
  while (b > 0) {
    if (b % 2 == 1) res = (res + a) % m;
    a = (a * 2) % m;
    b /= 2;
  }
  return res % m;
}
long long expMod(long long a, long long b, long long m = 1000000007) {
  long long x = 1, y = a;
  while (b) {
    if (b % 2 == 1) {
      x = mul(x, y, m);
    }
    y = mul(y, y, m);
    b = b / 2;
  }
  return x;
}
long long pw(long long a, long long b) {
  long long c = 1, m = a;
  while (b) {
    if (b & 1) c = (c * m);
    m = (m * m);
    b /= 2;
  }
  return c;
}
long long pwmd(long long a, long long b) {
  long long c = 1, m = a;
  while (b) {
    if (b & 1) c = (c * m) % 1000000007;
    m = (m * m) % 1000000007;
    b /= 2;
  }
  return c;
}
long long mmi(long long a) {
  long long inverse = pwmd(a, 1000000007 - 2);
  return inverse;
}
int my_log(long long n, int b) {
  long long i = 1;
  int ans = 0;
  while (1) {
    if (i > n) {
      ans--;
      break;
    }
    if (i == n) break;
    i *= b;
    ans++;
  }
  return ans;
}
inline bool kthbit(long long n, int k) { return (n >> k) & 1; }
inline long long setkthbit(long long n, int k) { return n | (1ll << k); }
inline long long unsetkthbit(long long n, int k) { return n & ~(1ll << k); }
inline long long flipkthbit(long long n, int k) { return n ^ (1ll << k); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long T;
  cin >> T;
  while (T--) {
    long long a, b, c, d, x, y, x1, y1, x2, y2;
    cin >> a >> b >> c >> d;
    cin >> x >> y >> x1 >> y1 >> x2 >> y2;
    x += b - a;
    y += d - c;
    int ok = 1;
    long long second = abs(x1 - x2);
    long long uu = abs(y1 - y2);
    if (second == 0 && uu == 0)
      ok = 0;
    else if (second == 0 && b == a && b > 0)
      ok = 0;
    else if (uu == 0 && c == d && c > 0)
      ok = 0;
    else if (!(x >= min(x1, x2) && x <= max(x1, x2) && y >= min(y1, y2) &&
               y <= max(y1, y2)))
      ok = 0;
    if (ok == 1) {
      cout << "YES"
           << "\n";
    } else {
      cout << "NO"
           << "\n";
    }
  }
}